"Throwing" in 3D

I wouldn't go so far as to actually implement a true physics calculation. Half a sine should do the trick for a realistic enough looking curve.

The general form of this problem is called "boundary value problem". The most primitive method of solving these problems is called the "shooting method". You try some values, look where you land and pick better start values till you hit close enough to your target.

What kind of physics do you want to consider? If you only want gravity it is easy:
1. reduce it to a 2d problem
2. pick some value for vx and solve for vy.
x_target=x0+vx*t
y_target=y0+vy*t-1/2*g*t*t

The sine method sounds good.

Yes, I only want to take in to account gravity. Enough so that the item has time to traverse the required distance before it lands.

A sine looks quite similar to the correct curve, but the correct one is almost as easy. I would use the correct one posted above.

Let's do a little physics.

Given starting position x₀ and y₀, and starting velocity v_0x and v_0y, we can calculate the final position by solving an constant-a equation (where we used the simplified model of gravity as the only accelerative force) in the y direction for time, then solving a constant-v equation in the x direction for x.
y_f = y₀ + v_0yt + 0.5at²
x_f = x₀ + v_0xt

a is the gravitational acceleration constant for your game (in real life, -9.8 m/s²)

So if you know the final position you want, x_f, just pick an arbitrary value for v₀ (representative of the unit's throwing power) and use v₀² = v_0x² + v_0y²

I'll leave all the dirty algebra up to you :D (let me know if you don't understand this)

Hope that helps,
Twilight Dragon

x=t*v_x+x₀
y=y₀+v_y0*t-g*t²/2
where t should be computed as t-t₀

x₀ horizontal initial position
v_x horizontal speed
y₀ vertical initial position
v_y0 ascendent initial speed
g=9.8*m/s² (convert in whatever)
t starts at the moment of throwing

EDIT:smbd.beat me to it

Thanks, guys. I'll look in to your suggestions.

Cheers.